# Prove that tan(a+b)=tana+tanb/1-tana*tanb

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### 2 Answers

First, you need to write this expression correctly, using brackets:

tan(a+b) = (tan a + tan b)/(1 - tan a*tan b)

We'll prove this identity, using the information that tangen function is a ratio:

tan (a+b) = sin (a+b)/cos (a+b)

We'll use the following identities:

sin (a+b) = sin a*cos b + sin b*cos a

cos (a+b) = cos a*cos b - sin a*sin b

tan (a+b) = (sin a*cos b + sin b*cos a)/(cos a*cos b - sin a*sin b)

We'll force factor cos a*cos b, both numerator and denominator, creating the tangent functions within brackets:

tan (a+b) = cos a*cos b( tan a + tan b)/cos a*cos b(1 - tan a*tan b)

We'll simplify and we'll get:

tan (a+b) = (tan a + tan b)/(1 - tan a*tan b)

**Therefore, the identity tan (a+b) = (tan a + tan b)/(1 - tan a*tan b) is verified.**

L:H:S = tan(a+b)

= sin(a+b)/cos(a+b)

= (sin a.cos b+cos a.sin b)/(cos a.cos b-sin a.sin b)

devide the numerator and denominator by cos a.cos b

= (sin a/cos a + sin b/cos b) / (1-sin a.sin b/cos a.cos b)

= (tan a + tan b)/(1 - tan a.tan b)

= R:H:S